Archimedean spiral in cartesian coordinates












0












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I know that the Archimedean spiral can be represented using the polar coordinate system very easily. But I was wondering if it can be represented using the Cartesian coordinate system, and if so what is the function?






geometry coordinate-systems polar-coordinates







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I know that the Archimedean spiral can be represented using the polar coordinate system very easily. But I was wondering if it can be represented using the Cartesian coordinate system, and if so what is the function?






    geometry coordinate-systems polar-coordinates







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I know that the Archimedean spiral can be represented using the polar coordinate system very easily. But I was wondering if it can be represented using the Cartesian coordinate system, and if so what is the function?






      geometry coordinate-systems polar-coordinates







      share|cite|improve this question











      $endgroup$




      I know that the Archimedean spiral can be represented using the polar coordinate system very easily. But I was wondering if it can be represented using the Cartesian coordinate system, and if so what is the function?






      geometry coordinate-systems polar-coordinates




      geometry coordinate-systems polar-coordinates






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 31 '18 at 0:34









      Jean Marie

      29.1k42050




      29.1k42050










      asked Jan 26 '18 at 18:01









      Dimitrios PilitsisDimitrios Pilitsis

      33




      33






















          1 Answer
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          $begingroup$

          Here is a solution for a double Archimedean spiral (see figure below).



          Let us consider the simplest Archimedean spiral with polar equation:



          $$tag{1}r=theta.$$



          Using the following formulas:



          $$tag{2}begin{cases}r^2=x^2+y^2\tan{theta}=tfrac{y}{x}\end{cases},$$



          (1) can be transformed into the following implicit cartesian equation:



          $$tag{3}arctan(tfrac{y}{x})=sqrt{x^2+y^2} (x neq 0).$$



          Taking $tan$ on both sides gives the solution:




          $$tag{4}y= x tansqrt{x^2+y^2}.$$




          Remarks:



          1) Note that, by multiplying by $x$, restriction $x neq 0$ is no longer needed. This shouldn't appear as magics : on the contrary, condition $x neq 0$ in (3) was depriving the curve from points $(0,(4k+1)tfrac{pi}{2}), k in mathbb{Z},$ that have been returned to their owner under the form (4).



          2) In fact, equation (4) defines a double Archimedean spiral (changing $(x,y)$ into $(-x,-y)$ doesn't change this equation). See picture below where the red curve is the Archimedean spiral, strictly speaking, and the magenta curve is its copy through a central symmetry.



          3) From (4), it doesn't look possible to extract explicit cartesian equations $y=f_n(x)$ (there would be of course an infinite number of such equations).



          4) (4) cannot be transformed into a polynomial implicit equation $P(x,y)=0$. A simple reason : any straight line intersects an Archimedean spiral in an infinite number of points, which is impossible for a polynomial equation.



          5) It is easy from there to obtain a cartesian equation analogous to (4) for the general Archimedean spirals $r=atheta.$



          6) An old identical question/answer: (http://mathforum.org/kb/message.jspa?messageID=685890) with a funny comment at the end.



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ Nominal Animal You are right. I will change it.
            $endgroup$
            – Jean Marie
            Jan 26 '18 at 19:52










          • $begingroup$
            @Nominal Animal About the absence of coefficient $b$, it is a choice : there must remain some work to do...
            $endgroup$
            – Jean Marie
            Jan 26 '18 at 19:53










          • $begingroup$
            I fully agree. I think that in this form (with the fourth point showing the way forward without showing the result), this answer is excellent.
            $endgroup$
            – Nominal Animal
            Jan 26 '18 at 20:16











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Here is a solution for a double Archimedean spiral (see figure below).



          Let us consider the simplest Archimedean spiral with polar equation:



          $$tag{1}r=theta.$$



          Using the following formulas:



          $$tag{2}begin{cases}r^2=x^2+y^2\tan{theta}=tfrac{y}{x}\end{cases},$$



          (1) can be transformed into the following implicit cartesian equation:



          $$tag{3}arctan(tfrac{y}{x})=sqrt{x^2+y^2} (x neq 0).$$



          Taking $tan$ on both sides gives the solution:




          $$tag{4}y= x tansqrt{x^2+y^2}.$$




          Remarks:



          1) Note that, by multiplying by $x$, restriction $x neq 0$ is no longer needed. This shouldn't appear as magics : on the contrary, condition $x neq 0$ in (3) was depriving the curve from points $(0,(4k+1)tfrac{pi}{2}), k in mathbb{Z},$ that have been returned to their owner under the form (4).



          2) In fact, equation (4) defines a double Archimedean spiral (changing $(x,y)$ into $(-x,-y)$ doesn't change this equation). See picture below where the red curve is the Archimedean spiral, strictly speaking, and the magenta curve is its copy through a central symmetry.



          3) From (4), it doesn't look possible to extract explicit cartesian equations $y=f_n(x)$ (there would be of course an infinite number of such equations).



          4) (4) cannot be transformed into a polynomial implicit equation $P(x,y)=0$. A simple reason : any straight line intersects an Archimedean spiral in an infinite number of points, which is impossible for a polynomial equation.



          5) It is easy from there to obtain a cartesian equation analogous to (4) for the general Archimedean spirals $r=atheta.$



          6) An old identical question/answer: (http://mathforum.org/kb/message.jspa?messageID=685890) with a funny comment at the end.



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ Nominal Animal You are right. I will change it.
            $endgroup$
            – Jean Marie
            Jan 26 '18 at 19:52










          • $begingroup$
            @Nominal Animal About the absence of coefficient $b$, it is a choice : there must remain some work to do...
            $endgroup$
            – Jean Marie
            Jan 26 '18 at 19:53










          • $begingroup$
            I fully agree. I think that in this form (with the fourth point showing the way forward without showing the result), this answer is excellent.
            $endgroup$
            – Nominal Animal
            Jan 26 '18 at 20:16
















          2












          $begingroup$

          Here is a solution for a double Archimedean spiral (see figure below).



          Let us consider the simplest Archimedean spiral with polar equation:



          $$tag{1}r=theta.$$



          Using the following formulas:



          $$tag{2}begin{cases}r^2=x^2+y^2\tan{theta}=tfrac{y}{x}\end{cases},$$



          (1) can be transformed into the following implicit cartesian equation:



          $$tag{3}arctan(tfrac{y}{x})=sqrt{x^2+y^2} (x neq 0).$$



          Taking $tan$ on both sides gives the solution:




          $$tag{4}y= x tansqrt{x^2+y^2}.$$




          Remarks:



          1) Note that, by multiplying by $x$, restriction $x neq 0$ is no longer needed. This shouldn't appear as magics : on the contrary, condition $x neq 0$ in (3) was depriving the curve from points $(0,(4k+1)tfrac{pi}{2}), k in mathbb{Z},$ that have been returned to their owner under the form (4).



          2) In fact, equation (4) defines a double Archimedean spiral (changing $(x,y)$ into $(-x,-y)$ doesn't change this equation). See picture below where the red curve is the Archimedean spiral, strictly speaking, and the magenta curve is its copy through a central symmetry.



          3) From (4), it doesn't look possible to extract explicit cartesian equations $y=f_n(x)$ (there would be of course an infinite number of such equations).



          4) (4) cannot be transformed into a polynomial implicit equation $P(x,y)=0$. A simple reason : any straight line intersects an Archimedean spiral in an infinite number of points, which is impossible for a polynomial equation.



          5) It is easy from there to obtain a cartesian equation analogous to (4) for the general Archimedean spirals $r=atheta.$



          6) An old identical question/answer: (http://mathforum.org/kb/message.jspa?messageID=685890) with a funny comment at the end.



          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ Nominal Animal You are right. I will change it.
            $endgroup$
            – Jean Marie
            Jan 26 '18 at 19:52










          • $begingroup$
            @Nominal Animal About the absence of coefficient $b$, it is a choice : there must remain some work to do...
            $endgroup$
            – Jean Marie
            Jan 26 '18 at 19:53










          • $begingroup$
            I fully agree. I think that in this form (with the fourth point showing the way forward without showing the result), this answer is excellent.
            $endgroup$
            – Nominal Animal
            Jan 26 '18 at 20:16














          2












          2








          2





          $begingroup$

          Here is a solution for a double Archimedean spiral (see figure below).



          Let us consider the simplest Archimedean spiral with polar equation:



          $$tag{1}r=theta.$$



          Using the following formulas:



          $$tag{2}begin{cases}r^2=x^2+y^2\tan{theta}=tfrac{y}{x}\end{cases},$$



          (1) can be transformed into the following implicit cartesian equation:



          $$tag{3}arctan(tfrac{y}{x})=sqrt{x^2+y^2} (x neq 0).$$



          Taking $tan$ on both sides gives the solution:




          $$tag{4}y= x tansqrt{x^2+y^2}.$$




          Remarks:



          1) Note that, by multiplying by $x$, restriction $x neq 0$ is no longer needed. This shouldn't appear as magics : on the contrary, condition $x neq 0$ in (3) was depriving the curve from points $(0,(4k+1)tfrac{pi}{2}), k in mathbb{Z},$ that have been returned to their owner under the form (4).



          2) In fact, equation (4) defines a double Archimedean spiral (changing $(x,y)$ into $(-x,-y)$ doesn't change this equation). See picture below where the red curve is the Archimedean spiral, strictly speaking, and the magenta curve is its copy through a central symmetry.



          3) From (4), it doesn't look possible to extract explicit cartesian equations $y=f_n(x)$ (there would be of course an infinite number of such equations).



          4) (4) cannot be transformed into a polynomial implicit equation $P(x,y)=0$. A simple reason : any straight line intersects an Archimedean spiral in an infinite number of points, which is impossible for a polynomial equation.



          5) It is easy from there to obtain a cartesian equation analogous to (4) for the general Archimedean spirals $r=atheta.$



          6) An old identical question/answer: (http://mathforum.org/kb/message.jspa?messageID=685890) with a funny comment at the end.



          enter image description here






          share|cite|improve this answer











          $endgroup$



          Here is a solution for a double Archimedean spiral (see figure below).



          Let us consider the simplest Archimedean spiral with polar equation:



          $$tag{1}r=theta.$$



          Using the following formulas:



          $$tag{2}begin{cases}r^2=x^2+y^2\tan{theta}=tfrac{y}{x}\end{cases},$$



          (1) can be transformed into the following implicit cartesian equation:



          $$tag{3}arctan(tfrac{y}{x})=sqrt{x^2+y^2} (x neq 0).$$



          Taking $tan$ on both sides gives the solution:




          $$tag{4}y= x tansqrt{x^2+y^2}.$$




          Remarks:



          1) Note that, by multiplying by $x$, restriction $x neq 0$ is no longer needed. This shouldn't appear as magics : on the contrary, condition $x neq 0$ in (3) was depriving the curve from points $(0,(4k+1)tfrac{pi}{2}), k in mathbb{Z},$ that have been returned to their owner under the form (4).



          2) In fact, equation (4) defines a double Archimedean spiral (changing $(x,y)$ into $(-x,-y)$ doesn't change this equation). See picture below where the red curve is the Archimedean spiral, strictly speaking, and the magenta curve is its copy through a central symmetry.



          3) From (4), it doesn't look possible to extract explicit cartesian equations $y=f_n(x)$ (there would be of course an infinite number of such equations).



          4) (4) cannot be transformed into a polynomial implicit equation $P(x,y)=0$. A simple reason : any straight line intersects an Archimedean spiral in an infinite number of points, which is impossible for a polynomial equation.



          5) It is easy from there to obtain a cartesian equation analogous to (4) for the general Archimedean spirals $r=atheta.$



          6) An old identical question/answer: (http://mathforum.org/kb/message.jspa?messageID=685890) with a funny comment at the end.



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 27 '18 at 11:19

























          answered Jan 26 '18 at 18:24









          Jean MarieJean Marie

          29.1k42050




          29.1k42050












          • $begingroup$
            @ Nominal Animal You are right. I will change it.
            $endgroup$
            – Jean Marie
            Jan 26 '18 at 19:52










          • $begingroup$
            @Nominal Animal About the absence of coefficient $b$, it is a choice : there must remain some work to do...
            $endgroup$
            – Jean Marie
            Jan 26 '18 at 19:53










          • $begingroup$
            I fully agree. I think that in this form (with the fourth point showing the way forward without showing the result), this answer is excellent.
            $endgroup$
            – Nominal Animal
            Jan 26 '18 at 20:16


















          • $begingroup$
            @ Nominal Animal You are right. I will change it.
            $endgroup$
            – Jean Marie
            Jan 26 '18 at 19:52










          • $begingroup$
            @Nominal Animal About the absence of coefficient $b$, it is a choice : there must remain some work to do...
            $endgroup$
            – Jean Marie
            Jan 26 '18 at 19:53










          • $begingroup$
            I fully agree. I think that in this form (with the fourth point showing the way forward without showing the result), this answer is excellent.
            $endgroup$
            – Nominal Animal
            Jan 26 '18 at 20:16
















          $begingroup$
          @ Nominal Animal You are right. I will change it.
          $endgroup$
          – Jean Marie
          Jan 26 '18 at 19:52




          $begingroup$
          @ Nominal Animal You are right. I will change it.
          $endgroup$
          – Jean Marie
          Jan 26 '18 at 19:52












          $begingroup$
          @Nominal Animal About the absence of coefficient $b$, it is a choice : there must remain some work to do...
          $endgroup$
          – Jean Marie
          Jan 26 '18 at 19:53




          $begingroup$
          @Nominal Animal About the absence of coefficient $b$, it is a choice : there must remain some work to do...
          $endgroup$
          – Jean Marie
          Jan 26 '18 at 19:53












          $begingroup$
          I fully agree. I think that in this form (with the fourth point showing the way forward without showing the result), this answer is excellent.
          $endgroup$
          – Nominal Animal
          Jan 26 '18 at 20:16




          $begingroup$
          I fully agree. I think that in this form (with the fourth point showing the way forward without showing the result), this answer is excellent.
          $endgroup$
          – Nominal Animal
          Jan 26 '18 at 20:16


















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